# Distance and Midpoint Formulas

The distance between two points and midpoint of a segment formulas are presented along with examples, questions , including detailed solutions .

## Distance Formula

Let us first define the distance $d$ between two points $A$ and $B$ whose coordinates are respectively $a$ and $b$ on a number line as $d = |a - b| = |b - a|$

Note that because of the absolute value, the distance between points is always positive or equal to zero.

We now review the
Pythagorean theorem : given a right triangle, as shown in figure 2 below, the hypotenuse (side opposite the right angle) and the two sides are related by: $h^2 = x^2 + y^2 \quad \text{or} \quad h = \sqrt {x^2 + y^2}$
We now use the distance between points on a number line and the Pythagorean theorem to the right triangle in figure 3 to write the formula for the distance between any two points on a plane.
Given two points $A = (a_x,a_y)$ and $B = (b_x,b_y)$, the formula that gives the distance $d$ between the two points $A$ and $B$, or the length of the line segment $AB$, is given by $\large \color{red}{d = \sqrt { (b_x - a_x)^2 + (b_y - a_y)^2} \qquad (I) }$

Note that   $|b_x - a_x|^2 = (b_x - a_x)^2$   and   $| b_y - a_y |^2 = (b_y - a_y)^2$

Example
Find the distance $d$ between the points $A = (3,2)$ and $B = (-3,-6)$
Solution
According to the formula (I), the distance $d$ is given by
$d = \sqrt{ (-3 - 3)^2 + (-6 - 2)^2 } = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36+64} = \sqrt {100} = 10$

## Midpoint Definition and Formula

The midpoint $M$ of the line segment $A B$ is the point on the line segment $A B$ such that the lengths of the segments $MA$ and $MB$ are equal. (See figure 4 below).
Given two points $A = (a_x,a_y)$ and $B = (b_x,b_y)$, the formula that give the
coordinates of the midpoint $M$ of the segment $AB$ is given by $\large \color{red}{ M = \left(\dfrac{a_x+b_x}{2} \; , \; \dfrac{a_y+b_y}{2} \right) } \qquad (II)$
Note that the $x$ and $y$ coordinates of $M$ are given by the average of the $x$ and $y$ coordinates of the points $A$ and $B$ respectively.

Example
Find the midpoint of the line segment $AB$ given that $A = (5,2)$ and $B = (-7,-10)$
Solution
According to the formula (II), the coordinates of the midpoint $M$ are given by
$M = \left( \dfrac{5+(-7)}{2} , \dfrac{2+(-10)}{2} \right) = \left( \dfrac{-2}{2} , \dfrac{-8}{2} \right) = ( -1 , -4 )$

## Questions

### Part A

Find the distance between each pair of points in the number lines shown below.

### Part B

Find the distance between the given pair of points.

1. $A = (0,0) \; , \; B = (-4,3)$
2. $C = (-2,-2) \; , \; D = (6,4)$
3. $E = (4,7) \; , \; F = (-4,7)$

### Part C

Find the midpoint between the segments defined by pair of points.

1. $A = (-8,0) \; , \; B = (-4,4)$
2. $C = (-3,-4) \; , \; D = (8,3)$
3. $E = (4,7) \; , \; F = (-4,7)$

### Part D

Given the points $A = (x_0,y_0)$ and $B = (-4,8)$. Find the coordinates of point $A$ if the midpoint of the segment $AB$ is the point $M = (-2,0)$.

### Part E

Given the points $A = (3 x_0, -y_0)$ and $B = ( x_0, 4 y_0)$. Find the coordinates of $A$ and $B$ if the midpoint of the segment $AB$ is the point $M = (2,6)$.

### Part F

1. Find the midpoint $M$ of the segment defined by the points $A = (-4,1)$ and $B = (2,5)$
2. Find the lengths of the segments $MA$ and $MB$. Are they equal?
3. Find the length of the line segment $AB$
4. Show that the length of the line segment $MA$ is half the length of the line segment $AB$.

## Solutions to the Above Questions

### Part A

The formula for the distance between two points on a number line is given above.

1. The coordinate of point $A$ is equal to $-2$ and the coordinate of point $B$ is equal to $9$; hence the distance $d_1$ between $A$ and $B$ is given by
$d_1 = | - 2 - 9 | = |-11| = 11$        or        $d_1 = | 9 - (-2) | = |11| = 11$
2. The coordinate of point $C$ is equal to $- 8$ and the coordinate of point $D$ is equal to $- 3$; hence the distance $d_2$ between $C$ and $D$ is given by
$d_2 = | -8 - (-3) | = |-8 + 3| = |-5| = 5$        or        $d_2 = | -3 - (-8) | = |-3 + 8| = |5| = 5$
3. The coordinate of point $E$ is equal to $3$ and the coordinate of point $F$ is equal to $10$; hence the distance $d_3$ between $E$ and $F$ is given by
$d_3 = | 3 - 10 | = |-7| = 7$        or        $d = | 10 - 3 | = |7| = 7$

### Part B

The formula for the distance between two points on a plane is given above.

1. $d(AB) = \sqrt { (-4 - 0)^2 + (3-0)^2} = \sqrt { (-4)^2 + (3)^2} = \sqrt { 16 + 9} = \sqrt{25} = 5$
2. $d(CD) = \sqrt { (6 - (-2))^2 + (4-(-2))^2} = \sqrt { (6+2)^2 + (4+2)^2} = \sqrt { 8^2 + 6^2 } = \sqrt{100} = 10$
3. $d (EF) = \sqrt { (-4 - 4)^2 + (7-7)^2} = \sqrt { (-8)^2 + 0 } = \sqrt { (-8)^2 } = \sqrt { 64 } = 8$

### Part C

The formula for the midpoint is given above.

1. $M_{AB} = \left(\dfrac{-8+(-4)}{2} , \dfrac{0+4}{2}\right) = \left(\dfrac{-12}{2} , \dfrac{4}{2}\right) = (-6 , 2)$
2. $M_{CD} = \left(\dfrac{-3+8}{2} , \dfrac{-4+3}{2}\right) = (5/2 , -1/2)$
3. $M_{EF} = \left(\dfrac{4+(-4)}{2} , \dfrac{7+7}{2}\right) = (0 , 7)$

### Part D

Given the points $A = (x_0,y_0)$ and $B = (-4,8)$, the use of the formula of the midpoint $M$ of $A B$ gives: $\quad M = \left( \dfrac{x_0 + (-4)}{2} , \dfrac{y_0 + 8}{2} \right)$
The coordinates of the midpoint are given as $(-2,0)$
We write that the coordinates of the midpoint are equal and therefore obtain the equations: $\dfrac{x_0 + (-4)}{2} = - 2$     (I)      and    $\dfrac{y_0 + 8}{2} = 0$     (II)
Multiply both sides of equation (I) above by $2$: $\quad 2 \times \dfrac{x_0 + (-4)}{2} = 2 \times (- 2)$
Simplify: $\quad x_0 - 4 = -4$
Solve for $x_0$ to obtain $x_0 = 0$
Multiply both sides of equation (II) above by $2$:$\quad 2 \times \dfrac{y_0 + 8}{2} = 2 \times (0)$
Simplify: $\quad y_0 + 8 = 0$
Solve for $y_0$ to obtain $y_0 = -8$
Since of $A = (x_0,y_0)$, the coordinates of $A$ are given by: $\quad A = (0,-8)$

### Part E

Given the points $A = (3 x_0, -y_0)$ and $B = ( x_0, 4 y_0)$, the midpoint formula gives: $\quad M = \left( \dfrac{3 x_0 + x_0}{2} , \dfrac{-y_0 + 4 y_0}{2} \right)$
Group like terms in the above to obtain: $\quad M = \left( \dfrac{4 x_0}{2} , \dfrac{3 y_0}{2} \right)$
The coordinates of the midpoint $M$ are known as $(2,6)$
Hence the equality of the coordinates gives the equations: $\dfrac{4 x_0}{2} = 2$     (I)      and      $\dfrac{3 y_0}{2} = 6$    (II)
Multiply both sides of equation (I) above by $2$: $\quad 2 \times \dfrac{4 x_0}{2} = 2 \times 2$
Simplify: $\quad 4 x_0 = 4$
Solve for $x_0$ to obtain $x_0 = 1$
Multiply both sides of equation (II) above by $2$:$\quad 2 \times \dfrac{3 y_0}{2} = 2 \times 6$
Simplify: $\quad 3 y_0 = 12$
Solve for $y_0$ to obtain $y_0 = 4$
Since of $A = (3x_0,-y_0)$, the coordinates of $A$ are given by: $\quad A = (3,-4)$
The coordinates of $B = ( x_0, 4 y_0)$ are given by: $\quad B = (1,16)$

### Part F

1. Use the midpoint formula: $\quad M = \left(\dfrac{-4+2}{2} , \dfrac{1+5}{2} \right) = ( -1 , 3)$

The length of a line segment whose endpoints are $A$ and $B$ is equal to the distance between $A$ and $B$ whose formula is given above.
2. The length of segment $MA$ is given by the distance formula: $\quad d( MA ) = \sqrt { (-4 - (-1) )^2 + (1 - 3 )^2 } = \sqrt { (-3)^2 + (-2 )^2 } = \sqrt {9+4} = \sqrt {13}$
The length of segment $MB$ is given by: $\quad d( MB ) = \sqrt { (2 - (-1) )^2 + (5 - 3 )^2 } = \sqrt { (3)^2 + (2 )^2 } = \sqrt {9+4} = \sqrt {13}$
The lengths $d(MA)$ and $d(MB)$ of the segments $MA$ and $MB$ respectively are equal as expected because of the definition of the midpoint given above.

3. The length $d(AB)$ is given by: $\quad d(AB) = \sqrt{ (2-(-4))^2 + (5-1)^2} = \sqrt{ 6^2 + 4^2} = \sqrt {52} = \sqrt {4 \times 13} = 2 \sqrt {13}$

4. $\dfrac{1}{2} \times d(AB) = \dfrac{1}{2} \times 2 \sqrt {13} = \sqrt {13} = d(MA)$.
Again the above is expected from the definition of the midpoint given above.

## More References and Links to Triangles

plotting points in rectangular coordinate system
Pythagorean Theorem and Problems with Solutions
Geometry Tutorials and Problems
The Four Pillars of Geometry - John Stillwell - Springer; 2005th edition (Aug. 9 2005) - ISBN-10 : 0387255303
Geometry: A Comprehensive Course - Daniel Pedoe - Dover Publications - 2013 - ISBN: 9780486131733
Geometry: with Geometry Explorer - Michael Hvidsten - McGraw Hill - 2006 - ISBN: 0-07-294863-9